Question

a, giải phương trình: \(x^2+2=2\sqrt{x^2+1}\)

b, Tìm các số nguyên x,y thỏa mãn: \(x^2+x+2y^2+y=2xy^2+xy+3\)

Answer 1

a, \(x^2+2=2\sqrt{x^2+1}\)

\(\Rightarrow x^2+1-2\sqrt{x^2+1}+1=0\)

\(\Rightarrow\left(\sqrt{x^2+1}-1\right)^2=0\)

\(\Rightarrow\sqrt{x^2+1}-1=0\)\(\Rightarrow x^2+1=1\Rightarrow x=0\)

b,\(x^2+x+2y^2+y=2xy^2+xy+3\)

\(\Rightarrow2xy^2+xy-x^2-x-2y^2-y+3=0\)

\(\Rightarrow2y^2\left(x-1\right)+y\left(x-1\right)-x\left(x-1\right)-2\left(x-1\right)+1=0\)

\(\Rightarrow\left(x-1\right)\left(2y^2+y-x-2\right)=-1=1\cdot\left(-1\right)=\left(-1\right)\cdot1\)

đoạn sau bạn tự giái tiếp nhé

Answer 2

a) \(x^2+2=2\sqrt{x^2+1}\)

\(\Leftrightarrow\left(x^2+2\right)^2=\left(2\sqrt{x^2+1}\right)^2\)

\(\Leftrightarrow x^4+4x^2+4=4x^2+4\)

\(\Leftrightarrow x=0\)

Answer 3

a/ \(x^2+2=2\sqrt{x^3+1}\)

\(\Leftrightarrow\left(x^2+2\right)^2=4\left(x^3+1\right)\)

\(\Leftrightarrow\left(x-2\right)^2x^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)